For getting now the derivatived⁢yd⁢xdydx\displaystyle\frac{dy}{dx} in a pointP0subscriptP0P_{0} of the curve, we chose another point PPP of the curve. If the values of the parametre ttt corresponding these points are t0subscriptt0t_{0} and ttt, we thus have the points (x⁢(t0),y⁢(t0))xsubscriptt0ysubscriptt0(x(t_{0}),\,y(t_{0})) and (x⁢(t),y⁢(t))xtyt(x(t),\,y(t)) and the slope of the secant line through the points is the difference quotient

Let us assume that the functions (1) are differentiable when t=t0tsubscriptt0t=t_{0} and that x′⁢(t0)≠0superscriptxnormal-′subscriptt00x^{{\prime}}(t_{0})\neq 0. As we let t→t0normal-→tsubscriptt0t\to t_{0}, the left side of (2) tends to the derivative d⁢yd⁢xdydx\frac{dy}{dx} and the right side to the quotienty′⁢(t0)x′⁢(t0)superscriptynormal-′subscriptt0superscriptxnormal-′subscriptt0\frac{y^{{\prime}}(t_{0})}{x^{{\prime}}(t_{0})}. Accordingly we have the result